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Proof step 6 from proof prefix-fwd-in (
previous step
)
n ∈ st0, ∀ n. n ∈ st0 ∧ n ∈ st1 → st0[n].list ⊑ st1[n].list ∧ st0[n].ladrlist ⊑ st1[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st1[n].list = # st1[n].ladrlist, ∀ n0. n0 ∈ st0 → n0 ∈ st1 ⊦ n ∈ st1
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: n
Additional information
Used terms: n